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Chapter 24

Chapter 24. Transition Metals and Coordination Compounds. Contents. The Colors of Rubies and Emeralds Properties of Transition Metals Coordination Compounds Structure and Isomerization Bonding in Coordination Compounds Applications of Coordination Compounds.

abra-weaver
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Chapter 24

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  1. Chapter 24 Transition Metals and Coordination Compounds

  2. Contents • The Colors of Rubies and Emeralds • Properties of Transition Metals • Coordination Compounds • Structure and Isomerization • Bonding in Coordination Compounds • Applications of Coordination Compounds

  3. The Colors of Rubies and Emeralds About 1% of the Al3+ ions in Al2O3 are replaced by Cr3+. About 1% of the Al3+ ions in Be3Al2(SiO3)6 are replaced by Cr3+.

  4. Werner’s Theory of Coordination Chemistry The two compounds have very similar formulas but are very different in appearance because of the different chemical structures.

  5. Properties of Transition Metals • For the 4th period d-block elements:

  6. Electron configurations [noble gas]ns2(n-1)dx [noble gas]ns2(n-2)f14(n-1)dx x: 1~10

  7. First-Row Transition Metal Orbital Occupancy

  8. Example 24.1 ***** Write the ground state electron configuration for Zr. Solution [Kr] 5s24d2 Example 24.2 Write the ground state electron configuration for Co3+. Solution For Co [Ar] 3d7 4s2 For Co3+ [Ar] 3d6

  9. Coordination Compounds • Terminology • A complex consists of a central atom, which is usually a metal atom or ion, and attached groups called ligands. • If a complex carries a net electric charge, it is called a complex ion • For a substance consisting of one or more complexes is called a coordination compound • The total number of points at which a central atom or ion attaches ligands, called coordination number.

  10. Bonding in Complex • Coordinate covalent bond (dative bond): The covalent bonding between two atoms in which both electrons come from only one of the atoms (of the ligand). • Central metal atom (ion): Lewis acid, electron pair acceptor • Ligand: Lewis base, electron pair donor • Chelating agent (chelator): polydentate ligand, for example: • Ethylenediamine (en): bidentate • Oxalato (ox): bidentate • Ethylenediaminetetraacetato (EDTA): hexadentate

  11. Common Ligands

  12. Four Common Structures of Complex Ions

  13. Bidentate Ligands Coordinated to Co3+

  14. Hexadentate Ligands Coordinated to Co3+

  15. Example What are the coordination number and the oxidation number of the central atom in (a) [CoCl4(NH3)2]– and (b) [Ni(CO)4]? Solution: • Co: center atom, 6 ligands attached (4 Cl– and 2 NH3) coordination number: 6. Charge calculation: x – 4 + 0 = –1, x = +3 central ion is Co3+, oxidation number is +3. (b) coordination number: 4 central atom is Ni, oxidation number is 0.

  16. Naming Coordination Compounds ***** • Names and Formulas of Common Ligands:

  17. Naming Coordination Compounds • Naming complex cations: • Ligand first, metal (with oxidation number written in Roman numerals) after. • Name the ligands in alphabetical order (ignoring Greek numeric prefixes). • Designate the number of ligands in a complex with a Greek numeric prefix: di = 2, tri = 3, tetra = 4, hexa = 6. • If the ligand name itself includes a Greek numeric prefix, place parentheses around the ligand name, then prefix: bis = 2, tris = 3, tetrakis = 4

  18. Names of common metals when found in anionic complex Ions

  19. For a coordination compound: similar to naming ionic compound, cation first, anion after. • When writing a complex formula from name: • Center metal first, ligands after • Place the ligands in alphabetical order (ignoring Greek numeric prefixes)

  20. Examples of Naming Coordination Compounds *****

  21. Example ***** Write the formula for (a) triamminechlorodinitrito-O-platinum(IV) ion and (b) sodium hexanitrito-N-cobaltate(III). Solution: • Complex ion charge: +1 (1 Pt4+: +4, 3 NH3: 0, 1 Cl–: –1, 2 ONO–: –2), a complex cation Metal first, ligands after, order of alphabetical Ans: [Pt(NH3)3Cl(ONO)2]+ • Coordination compound: made up of Na+ cations and a complex anion. Complex anion charge: –3 (1 Co3+: +3, 6NO2–: –6) Cation first, anion after: Ans: Na3[Co(NO2)6].

  22. 4. Structure and Isomerization

  23. Structural isomers i) Linkage isomers: differ in the donor atoms through which the ligands are bonded, for example: [Co(NH3)5(NO2)]Cl2 [Co(NH3)5(ONO)]Cl2 ii) Coordination isomers: differ in the ligands that are attached to the central atom, for example: [Cr(NH3)5(SO4)]Cl [Co(NH3)5Cl]SO4

  24. 2) Stereoisomers • Geometric isomers a) Cis-trans isomers (MA2B2 and MA4B2 type)

  25. ***** b) Fac-mer isomers (MA3B3 type)

  26. Optical isomers (enantiomers): • Molecules that are nonsuperimposable (not identical) mirror images of one another, like right and left hands. • Each enantiomer rotates polarized light in opposite directions. • For example: Mirror image of each other Two nonsuperimposable (not identical) structures Ans: two optical isomers

  27. 5. Bonding in Coordination Compounds • Common Hybridization Schemes in Complex Ions

  28. Crystal Field Theory ***** • Attractions between a central atom (or ion) and its ligands are largely electrostatic. • Ligandsdistort the d-orbitals of the central atom, leading to a splitting of energy levels of those orbitals. • Splitting energy(Δ): The splitting of energy levels of d-orbitals which are caused by ligands distoration of those orbitals. • The spectrochemical series shows the relative abilities of ligands (Δ) to split the d-orbital energy levels: • Pairing Energy (P): Coulombic repulsion energy caused by having two electrons in same orbital.

  29. ***** • Schematic representation of d-level splitting: Δ Δ Usually Δ < P High Spin Usually Δ > P Low Spin Using Spectroche-mical series Spin: unpaired electron

  30. ***** • Crystal field theory can predict: • Magnetism • Paramagnetic, with spin (unpaired electron) • Diamagnetic, without spin • Magnetic strength • High spin, more unpaired electron (if Δsmall) • Low spin, less unpaired electron, (if Δ large) • Complex color: According to the energy level transition

  31. ***** Example: Two different ligands (H2O vs. CN–) Fe3+ complex Electron configuration: Fe: [Ar]3d64s2 Fe3+: [Ar]3d5 (five 3d electrons) Δ: CN– > H2O

  32. ***** Example How many unpaired electrons would you expect for the octahedral complex ion [CoF6]3–? Solution: Electron configuration: Co: [Ar]3d74s2 Co3+: [Ar]3d6 (six 3d electrons) F– ligand: Δ small Ans: 4 unpaired electrons

  33. ***** Example How many unpaired electrons would you expect to find in the tetrahedralcomplex ion [NiCl4]2–? Solution: Electron configuration: Ni: [Ar]3d84s2 Ni2+: [Ar]3d8 (Tetrahedral, usually Δ small, high spin) Ans: 2 unpaired electrons

  34. ***** Example How many unpaired electrons would you expect to find in the square planarcomplex ion [PtCl4]2–? Solution: Electron configuration: Pt: [Xe]5d96s1 Pt2+: [Xe]5d8 (Square planar, usually Δ large, low spin) Ans: no unpaired electrons a diamagnetic species

  35. Color In Complex Ions And Coordination Compounds • Many complex ions are colored because the energy differences between d orbitals match the energies of components of visible light. • Crystal field theory helps to explain the colors of complex ions. • Ions having the following electron configurations have no electron transitions in the energy range of visible light (colorless): • No electron in d orbital (d0-complex), e.g., Sc3+, Y3+, La3+ (noble-gas electron configuration) are colorless. • Electrons completely filled in d orbital (d10-complex), e.g., Zn2+, Cd2+, Hg2+, Cu+, and Ag+ (18 or 18+2 electron configuration) are colorless.

  36. The color of the transmitted light is the complementary color of the absorbed light. 700 Small Δ Low ν Long λ Wavelength (nm) Large Δ High ν Short λ 400

  37. Complex Ion Color and Crystal Field Strength • The colors of complex ions are due to electronic transitions between the split d sublevel orbitals. • The wavelength of maximum absorbance can be used to determine the size of the energy gap between the split d sublevel orbitals. • Ephoton = hn = hc/l = D

  38. ***** Example 24.8 The complex ion [Cu(NH3)6]2+ is blue in aqueous solution. Estimate the crystal field splitting energy (in kJ/mol) for this ion. Solution The color orange ranges from 580 to 650 nm, so you can estimate the average wavelength as 615 nm. E = hc/λ. Convert J/ion into kJ/mol.

  39. Extraction of metals from ores Silver and gold as cyanide complexes Nickel as Ni(CO)4(g) Use of chelating agents in heavy metal poisoning EDTA for Pb poisoning Chemical analysis Qualitative analysis for metal ions Blue = CoSCN+ Red = FeSCN2+ Ni2+ and Pd2+ form insoluble colored precipitates with dimethylglyoxime. 6. Applications of Coordination Compounds

  40. ***** Example Calculate the concentration of Ag+, in an aqueous solution prepared as 0.10 M AgNO3 and 3.0 M NH3. Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq) Kf = 1.6 x 107 Assume all shift right firstly Solution:

  41. End of Chapter 24

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